Mathematics · 2024
JEE Main · 9 April 2024, Shift 1 · Q6
Let f(x)=a x^3+b x^2+c x+41 be such that f(1)=40, f^′(1)=2 and f^′ ′(1)=4. Then a^2+b^2+c^2 is equal to:
Let $\displaystyle f(x)=a x^3+b x^2+c x+41$ be such that $\displaystyle f(1)=40, f^{\prime}(1)=2$ and $\displaystyle f^{\prime \prime}(1)=4$. Then $\displaystyle a^2+b^2+c^2$ is equal to:
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle 51$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.